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GWAE
GWAE
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August 17th, 2019 at 11:27:09 AM permalink
Quote: RS

Looks like no one is interested in this actual topic. Although, I think part of that might be because I worded it (partially) wrong.

If you win, then you would win $2, not $1. This would make the game +EV.

EG: If you bet $100, then you have a 99% chance to win $2 and a 1% chance to lose $100. Over 100 equally distributed games, you'd have won $198 and lost $100, for a profit of $98 on $10,000 in action, or a 0.98% edge. With a $1,000,000 wager, you'd win $1,999,998 and lose $1,000,000 for a net profit of $999,998 over $10,000,000,000,000 in action, for an edge of 0.0000000999998% edge. I might have messed up on the zeroes though (back off me chief). IOW: It's pretty much worth $1 per bet.

I'm sure everyone here would play the game for $100. But what if it was only offered for $1k, $10k, or $100k? Or better yet, would the most you'd be willing to bet be 1% of your current bankroll? 5%, 10%, 25%, 50%, or 100%?

Lol at Rigindux not knowing English



Isnt this basically the same thing as voting no safety except with no data? So it is bet 100 to win 1 or 10k to win 1 and so on. At current financial I could only risk 100, but I am not risking 100 for $1 even with 99% chance of win.
Expect the worst and you will never be disappointed. I AM NOT PART OF GWAE RADIO SHOW
RS
RS
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August 17th, 2019 at 1:10:18 PM permalink
Quote: GWAE

Quote: RS

Looks like no one is interested in this actual topic. Although, I think part of that might be because I worded it (partially) wrong.

If you win, then you would win $2, not $1. This would make the game +EV.

EG: If you bet $100, then you have a 99% chance to win $2 and a 1% chance to lose $100. Over 100 equally distributed games, you'd have won $198 and lost $100, for a profit of $98 on $10,000 in action, or a 0.98% edge. With a $1,000,000 wager, you'd win $1,999,998 and lose $1,000,000 for a net profit of $999,998 over $10,000,000,000,000 in action, for an edge of 0.0000000999998% edge. I might have messed up on the zeroes though (back off me chief). IOW: It's pretty much worth $1 per bet.

I'm sure everyone here would play the game for $100. But what if it was only offered for $1k, $10k, or $100k? Or better yet, would the most you'd be willing to bet be 1% of your current bankroll? 5%, 10%, 25%, 50%, or 100%?

Lol at Rigindux not knowing English



Isnt this basically the same thing as voting no safety except with no data? So it is bet 100 to win 1 or 10k to win 1 and so on. At current financial I could only risk 100, but I am not risking 100 for $1 even with 99% chance of win.


Yes, it's a bridge jumper as I think FleaStiff said earlier in the thread.

And it'd be to win $2, not $1.
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TomG
TomG
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Thanks for this post from:
RSRigondeaux
August 17th, 2019 at 2:30:22 PM permalink
Quote: RS

If you win, then you would win $2, not $1. This would make the game +EV



So no matter what I bet, the value is +$1? I would just reach into my pocket and bet whatever amount I pulled out. And if I could repeat that bet, I would keep doing so until I was rich.

With a minimum bet of $1,000, I would probably just keep betting $1,000 until all those $2 wins added up to a real nice net worth -- assuming it's like a slot machine with hundreds of decisions per hour. My rough estimates show up to 50% of bankroll should be ok (lets say 40% for being conservative), I would still prefer less. At a certain point it becomes almost exactly like a reverse powerball. Or more accurately, it is just booking powerball bets (only the jackpot never grows, but the winners don't have to split their prizes), which we know is a great way to make money.
RS
RS
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August 17th, 2019 at 8:42:14 PM permalink
Quote: TomG

So no matter what I bet, the value is +$1? I would just reach into my pocket and bet whatever amount I pulled out. And if I could repeat that bet, I would keep doing so until I was rich.

With a minimum bet of $1,000, I would probably just keep betting $1,000 until all those $2 wins added up to a real nice net worth -- assuming it's like a slot machine with hundreds of decisions per hour. My rough estimates show up to 50% of bankroll should be ok (lets say 40% for being conservative), I would still prefer less. At a certain point it becomes almost exactly like a reverse powerball. Or more accurately, it is just booking powerball bets (only the jackpot never grows, but the winners don't have to split their prizes), which we know is a great way to make money.


I came to the same conclusion, that the maximum (full kelly) bet would 50% of BR. Although it is kinda weird, since you normally want to bet the most you can, in this case, that isnít really applicable because betting more doesnít get you (that much) more EV.
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Ayecarumba
Ayecarumba
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August 17th, 2019 at 10:25:22 PM permalink
Thereís a point where a large enough wager makes your chance of losing, ďthe same as 0%Ē. How large would the wager need to be?

I ask because it was established some time ago in another thread that 0.99999.... = 1.
Simplicity is the ultimate sophistication - Leonardo da Vinci
RS
RS
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August 17th, 2019 at 11:03:44 PM permalink
Quote: Ayecarumba

Thereís a point where a large enough wager makes your chance of losing, ďthe same as 0%Ē. How large would the wager need to be?

I ask because it was established some time ago in another thread that 0.99999.... = 1.


0.999 has to repeat forever, which would mean the wager would have to be infinite.
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kubikulann
kubikulann
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August 18th, 2019 at 4:10:00 AM permalink
Quote: RS

I came to the same conclusion, that the maximum (full kelly) bet would 50% of BR. Although it is kinda weird, since you normally want to bet the most you can, in this case, that isnít really applicable because betting more doesnít get you (that much) more EV.

Kelly criterion is for long term growth. If horizon is finite, the formula is Max E[G^(1/n)], with G=resulting capital and n=number of plays. Hence for one try, it says Bet All when return is proportional to wager (w).

Édit: and positive! The original game had a negative EV.
The new one has (1)/w * (-w) + (w-1)/w * 2 = (w-2)/w = 1 - 2/w which is not proportional to w.
In Kelly parlance, EG = 1/w * (B-w) + (w-1)/w * (B+2) = (B-w+wB-B+2w-2)/w = (wB+w-2)/w = B+1-2/w.
Maximum for largest w as possible.

This result shows the weakness of Kelly play. It does not take into account the variance. Or equivalently, it ignores risk-aversion and loss-aversion. What human would risk all their capital to win 2$? They do the reverse: buy for 2$ a lottery ticket ; pay 2$ premium to insure against the loss of a capital.

RSís question is precisely : « what is your risk-aversion? »
His game is akin to Bernoulliís St-Petersburg bet.

- -

Furthermore, the Kelly criterion is based on a traditional bet, i.e. one with fixed probabilities. Here, probabilities are dependent on w. This voids all the mathematical development of the Kelly formula. (Because the derivative of the expectation is NOT anymore the expectation of the derivative.)

Development:
Max E[G^(1/n)] = (1/w){ (B-w)^(1/n) + (w-1)*(B+2)^(1/n) }

Derivative w.r.t. w = (-1/ww) {Ö} + (1/w) { -(1/n)(B-w)^(1/n-1) + (w-1)*(1/n)(B+2)^(1/n-1) + (B+2)^(1/n) }
= (1/w) * { (B-w)^(1/n) * [-1/w - (B-w)^(-1)/n] + (B+2)^(1/n) * [1/w + (w-1) (B+2)^(-1)/n] }
= (1/ww) { (B+2)^(1/n) - (B-w)^(1/n) } + (1/nw) { (w-1) (B+2)^(1/n)/(B+2) - (B-w)^(1/n)/(B-w) }
= 0

Limit for n-> inf
(1/ww) {epsilon} + (1/nw) { (w-1)/(B+2) - 1/(B-w) } = 0
<=> (w-1)/(B+2) = 1/(B-w)
<=> (w-1)(B-w) = (B+2) Quadratic in w: ww - (B+1) w - 2 = 0
<=> w = (B+1)/2 +/- sqroot[((B+1)/2)^2 + 2]
This outside the bounds 0<w<B so to find a maximum we look at the sign of (w-1)/(B+2) - 1/(B-w)
= sign of (w-1)(B-w) - (B+2) if w<B
= positives between the roots
So the maximum is reached by increasing w.

A long term risk-neutral Kelly bettor invests his whole bankroll on this game!
Ainít the Kelly criterion silly?😋
we all know this leads to assured ruin.
Last edited by: kubikulann on Aug 18, 2019
Reperiet qui quaesiverit
RS
RS
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August 18th, 2019 at 6:25:59 AM permalink
Quote: kubikulann

Kelly criterion is for long term growth. If horizon is finite, the formula is Max E[G^(1/n)], with G=resulting capital and n=number of plays. Hence for one try, it says Bet All when return is proportional to wager (w).

Édit: and positive! The original game had a negative EV.
The new one has (1)/w * (-w) + (w-1)/w * 2 = (w-2)/w = 1 - 2/w which is not proportional to w.
In Kelly parlance, EG = 1/w * (B-w) + (w-1)/w * (B+2) = (B-w+wB-B+2w-2)/w = (wB+w-2)/w = B+1-2/w.
Maximum for largest w as possible.

This result shows the weakness of Kelly play. It does not take into account the variance. Or equivalently, it ignores risk-aversion and loss-aversion. What human would risk all their capital to win 2$? They do the reverse: buy for 2$ a lottery ticket ; pay 2$ premium to insure against the loss of a capital.

RSís question is precisely : « what is your risk-aversion? »
His game is akin to Bernoulliís St-Petersburg bet.

- -

Furthermore, the Kelly criterion is based on a traditional bet, i.e. one with fixed probabilities. Here, probabilities are dependent on w. This voids all the mathematical development of the Kelly formula. (Because the derivative of the expectation is NOT anymore the expectation of the derivative.)
Development:
Max E[G^(1/n)] = (1/w){ (B-w)^(1/n) + (w-1)*(B+2)^(1/n) }
Derivative w.r.t. w = (-1/ww) {Ö} + (1/w) { -(1/n)(B-w)^(1/n-1) + (w-1)*(1/n)(B+2)^(1/n-1) + (B+2)^(1/n) }
= (1/w) * { (B-w)^(1/n) * [-1/w - (B-w)^(-1)/n] + (B+2)^(1/n) * [1/w + (w-1) (B+2)^(-1)/n] }
= (1/ww) { (B+2)^(1/n) - (B-w)^(1/n) } + (1/nw) { (w-1) (B+2)^(1/n)/(B+2) - (B-w)^(1/n)/(B-w) }
=


I don't understand all of that math. But wouldn't it be fair to say if someone asked you if you'd make a certain bet, you'd apply the kelly criterion, regardless of any other potential bets? Why does it matter if the probability is based on the size of the wager? Wouldn't it be the same as if you wanted to determine if taking a coin flip at +105 odds (1.05-to-1 payout) for $10k is within your risk tolerance? But in this case, you want to see if betting $100k to win $2 in a reverse lottery is within your risk tolerance.
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kubikulann
kubikulann
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August 18th, 2019 at 6:45:27 AM permalink
Quote: RS

I don't understand all of that math. But wouldn't it be fair to say if someone asked you if you'd make a certain bet, you'd apply the kelly criterion, regardless of any other potential bets?

No. As I said, the Kelly criterion is designed for optimizing the long term growth of a capital/bankroll in a repeated positive EV situation. And a situation where the outcomes are proportional to the wager.
If someone asks me whether Iíd make a bet, I donít think in long term (itís a one shot) and, more importantly, I am not risk-neutral. And here, the outcome (2$) is not proportional.
In this here game, Kelly betting leads to ruin in the long run.
Quote: RS

Why does it matter if the probability is based on the size of the wager? .

Because the demonstration of the formula requires to pass from the derivative of E(x) to E(derivative of x).
Say E(G(x))=af(x)+bg(x) where a and b are constants (probabilities).
Then E(G(x))í=afí(x)+bgí(x)=E(Gí(x))
But if E(G(x))=a(x)f(x)+b(x)g(x) , then itís derivative is not a(x)fí(x)+b(x)gí(x) =E(Gí(x)).

In other words, it assumes repetition of the same bet: same probability - yet here the probability depends on w, which depends on B, which varies along the way.
Last edited by: kubikulann on Aug 18, 2019
Reperiet qui quaesiverit
RS
RS
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August 18th, 2019 at 6:58:00 AM permalink
Quote: kubikulann

No. As I said, the Kelly criterion is designed for optimizing the long term growth of a capital/bankroll in a repeated positive EV situation. And a situation where the outcomes are proportional to the wager.
If someone asks me whether Iíd make a bet, I donít think in long term (itís a one shot) and, more importantly, I am not risk-neutral. And here, the outcome (2$) is not proportional. Because the demonstration of the formula requires to pass from the derivative of E(x) to E(derivative of x).
In other words, it assumes repetition of the same bet: same probability - yet here the probability depends on w, which depends on B, which varies along the way.


Okay, so if you can't apply the kelly criterion, then what's the solution? How do you figure out what is the maximum amount to wager, if offered at that amount? IOW, what size of wager brings too high of a risk?
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